Permutation is an arrangement of objects in a specific order. Order of arrangement of object is very important. The number of permutations on a set of n elements is given by n!. For example, there are 2! = 2*1 = 2 permutations of {1, 2}, namely {1, 2} and {2, 1}, and 3! = 3*2*1 = 6 permutations of {1, 2, 3}, namely {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2} and {3, 2, 1}.
Method 1 (Backtracking)
We can use the backtracking based recursive solution discussed here.
Method 2
The idea is to one by one extract all elements, place them at first position and recur for remaining list.
# Python function to print permutations of a given list def permutation(lst):
# If lst is empty then there are no permutations
if len (lst) = = 0 :
return []
# If there is only one element in lst then, only
# one permutation is possible
if len (lst) = = 1 :
return [lst]
# Find the permutations for lst if there are
# more than 1 characters
l = [] # empty list that will store current permutation
# Iterate the input(lst) and calculate the permutation
for i in range ( len (lst)):
m = lst[i]
# Extract lst[i] or m from the list. remLst is
# remaining list
remLst = lst[:i] + lst[i + 1 :]
# Generating all permutations where m is first
# element
for p in permutation(remLst):
l.append([m] + p)
return l
# Driver program to test above function data = list ( '123' )
for p in permutation(data):
print (p)
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Output:
['1', '2', '3'] ['1', '3', '2'] ['2', '1', '3'] ['2', '3', '1'] ['3', '1', '2'] ['3', '2', '1']
Method 3 (Direct Function)
We can do it by simply using the built-in permutation function in itertools library. It is the shortest technique to find the permutation.
from itertools import permutations
l = list (permutations( range ( 1 , 4 )))
print (l)
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Output:
[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)]